Nabla Operator

In 3D Cartesian coordinates,

\nabla =\frac{\partial}{\partial x}\hat{x}+\frac{\partial}{\partial y}\hat{y}+\frac{\partial}{\partial z}\hat{z}

To obtain its spherical form, we need to use the law of chain, by using:

{ \left\{ \begin{aligned} r&=\sqrt{x^2+y^2+z^2}\\ \theta &=\arccos \frac{z}{r}\\ \phi &=\arctan \frac{y}{x}\\ \end{aligned} \right. }

so that

\begin{aligned}\frac{\partial}{\partial x}&=\frac{\partial}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial}{\partial \theta}\frac{\partial \theta}{\partial x}+\frac{\partial}{\partial \phi}\frac{\partial \phi}{\partial x} \\ &=\frac{\partial}{\partial r}\frac{x}{r}+\frac{\partial}{\partial \theta}\frac{xz/r^3}{\sqrt{1-\left( z/r \right) ^2}}-\frac{\partial}{\partial \phi}\frac{y/x^2}{1+\left( y/x \right) ^2} \\ &=\frac{\partial}{\partial r}\sin \theta \cos \phi +\frac{\partial}{\partial \theta}\frac{\cos \theta \cos \phi}{r}-\frac{\partial}{\partial \phi}\frac{\sin \phi}{r\sin \theta}\end{aligned}

There’s one thing to be noticed: you shouldn’t directly use

{ \left\{ \begin{aligned} x&=r\sin \theta \cos \phi\\ y&=r\sin \theta \sin \phi\\ z&=r\cos \theta\\ \end{aligned} \right. }

and wrongly assume that \partial x/\partial r=1/(\partial r/\partial x).

Similarly,

{\frac{\partial}{\partial y}=\frac{\partial}{\partial r}\sin \theta \sin \phi +\frac{\partial}{\partial \theta}\frac{\cos \theta \sin \phi}{r}+\frac{\partial}{\partial \phi}\frac{\cos \phi}{r\sin \theta}}

{\frac{\partial}{\partial z}=\frac{\partial}{\partial r}\cos \theta -\frac{\partial}{\partial \theta}\frac{\sin \theta}{r}}

The notation here might be a bit confusing, because I didn’t notice that the forms above could be wrongly interpreted as differential operators operating on existing expressions, like: \partial/\partial \theta (\cos \theta) \rightarrow -\sin\theta; however, that’s not what I meant; here, it should be just a simple multiplication. You can see that in Equation 1, I finally thought of this and adjusted the ordering of expressions so that it’s much clearer.

We also need to transform the basis vectors:

{\left\{ \begin{aligned} \hat{x}&=\hat{r}\sin \theta \cos \phi +\hat{\theta}\cos \theta \cos \phi -\hat{\phi}\sin \phi\\ \hat{y}&=\hat{r}\sin \theta \sin \phi +\hat{\theta}\cos \theta \sin \phi +\hat{\phi}\cos \phi\\ \hat{z}&=\hat{r}\cos \theta -\hat{\theta}\sin \theta\\ \end{aligned} \right. }

Then:

\begin{aligned} \nabla =& \frac{\partial}{\partial x}\hat{x}+\frac{\partial}{\partial y}\hat{y}+\frac{\partial}{\partial z}\hat{z} \\ =&\left( \frac{\partial}{\partial r}\sin \theta \cos \phi +\frac{\partial}{\partial \theta}\frac{\cos \theta \cos \phi}{r}-\frac{\partial}{\partial \phi}\frac{\sin \phi}{r\sin \theta} \right) \left( \hat{r}\sin \theta \cos \phi +\hat{\theta}\cos \theta \cos \phi -\hat{\phi}\sin \phi \right) \\ &+ \left( \frac{\partial}{\partial r}\sin \theta \sin \phi +\frac{\partial}{\partial \theta}\frac{\cos \theta \sin \phi}{r}+ \frac{\partial}{\partial \phi}\frac{\cos \phi}{r\sin \theta} \right) \left( \hat{r}\sin \theta \sin \phi +\hat{\theta}\cos \theta \sin \phi +\hat{\phi}\cos \phi \right) \\ &+\left( \frac{\partial}{\partial r}\cos \theta -\frac{\partial}{\partial \theta}\frac{\sin \theta}{r} \right) \left( \hat{r}\cos \theta -\hat{\theta}\sin \theta \right) \\ =&\hat{r}\frac{\partial}{\partial r}+\hat{\theta}\frac{1}{r}\frac{\partial}{\partial \theta}+\hat{\phi}\frac{1}{r\sin \theta}\frac{\partial}{\partial \phi} \end{aligned} \tag{1}

For gradient \nabla \varphi, there’s no problem using Equation 1 directly; however, we can’t directly use it to calculate divergence \nabla \cdot \vec{E} and curl \nabla \times\vec{E}; that’s because spherical coordinates itself has non-0 divergence and curl. By using

{\begin{cases} \hat{r}=\hat{x}\cos \phi \sin \theta +\hat{y}\sin \phi \sin \theta +\hat{z}\cos \theta\\ \hat{\theta}=\hat{x}\cos \phi \cos \theta +\hat{y}\sin \phi \cos \theta -\hat{z}\sin \theta\\ \hat{\phi}=-\hat{x}\sin \phi +\hat{y}\cos \phi\\ \end{cases}}

we can obtain that

\left\{ \begin{aligned} \nabla \cdot \hat{r} &=\frac{\partial}{\partial x}\left( \frac{x}{r} \right) +\frac{\partial}{\partial y}\left( \frac{y}{r} \right) +\frac{\partial}{\partial z}\left( \frac{z}{r} \right) = \frac{2}{r} \\ \nabla \times \hat{r} &= 0 \\ ... \end{aligned} \right.

Divergence \nabla\cdot\vec{E}

First, we need to calculate the differential terms produced by basis vectors:

{\left\{ \begin{aligned} \frac{\partial \hat{r}}{\partial r}&=\frac{\partial \hat{\theta}}{\partial r}=\frac{\partial \hat{\phi}}{\partial r}=0\\ \frac{\partial \hat{r}}{\partial \theta}&=\hat{\theta}, \frac{\partial \hat{r}}{\partial \phi}=\hat{\phi}\sin \theta\\ \frac{\partial \hat{\theta}}{\partial \theta}&=-\hat{r}, \frac{\partial \hat{\theta}}{\partial \phi}=\hat{\phi}\cos \theta\\ \frac{\partial \hat{\phi}}{\partial \theta}&=0, \frac{\partial \hat{\phi}}{\partial \phi}=-\left( \hat{r}\sin \theta +\hat{\theta}\cos \theta \right)\\ \end{aligned} \right. }

Then:

\begin{aligned} \nabla \cdot \vec{E}&=\left( \hat{r}\frac{\partial}{\partial r}+\hat{\theta}\frac{1}{r}\frac{\partial}{\partial \theta}+\hat{\phi}\frac{1}{r\sin \theta}\frac{\partial}{\partial \phi} \right) \cdot \left( E_r\hat{r}+E_{\theta}\hat{\theta}+E_{\phi}\hat{\phi} \right) \\ &=\frac{\partial E_r}{\partial r}+\frac{1}{r}\frac{\partial E_{\theta}}{\partial \theta}+\frac{1}{r\sin \theta}\frac{\partial E_{\phi}}{\partial \phi}+\frac{2E_r}{r}+\frac{E_{\theta}}{r\tan \theta} \\ & =\frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2E_r \right) +\frac{1}{r\sin \theta}\frac{\partial}{\partial \theta}\left( E_{\theta}\sin \theta \right) +\frac{1}{r\sin \theta}\frac{\partial E_{\phi}}{\partial \phi} \end{aligned} \tag{2}

Curl \nabla\times\vec{E}

With the same principle, we can get:

\nabla \times\vec{E} = \frac{1}{r^2\sin\theta} \left| \begin{matrix} \hat{r} & r\hat{\theta} & r\sin\theta\hat{\phi} \\ \partial_r & \partial_\theta & \partial_\phi \\ E_r & rE_\theta & r\sin\theta E_\phi \end{matrix}\right| \tag{3}

Laplace Operator \nabla^2

After all these calculations,

\begin{aligned} \nabla ^2&=\nabla \cdot \left( \hat{r}\frac{\partial}{\partial r}+\hat{\theta}\frac{1}{r}\frac{\partial}{\partial \theta}+\hat{\phi}\frac{1}{r\sin \theta}\frac{\partial}{\partial \phi} \right) \\ & =\frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2\frac{\partial}{\partial r} \right) +\frac{1}{r^2\sin \theta}\frac{\partial}{\partial \theta}\left( \sin \theta \frac{\partial}{\partial \theta} \right) +\frac{1}{r^2\sin ^2\theta}\frac{\partial ^2}{\partial \phi ^2} \end{aligned} \tag{4}